Find $\lim_{x\to 4}\dfrac{3x}{(x-2)(x+2)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{4}$ (Choice B) B $\dfrac{3}{4}$ (Choice C) C $1$ (Choice D) D The limit doesn't exist
Solution: Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to 4}\dfrac{3x}{(x-2)(x+2)}&=\dfrac{3(4)}{(4-2)(4+2)} \\\\ &=\dfrac{12}{(2)(6)} \\\\ &=\dfrac{12}{12} \\\\ &=1 \end{aligned}$ We got a finite number. Since $\dfrac{3x}{(x-2)(x+2)}$ is continuous across its domain, we can determine that $\lim_{x\to 4}\dfrac{3x}{(x-2)(x+2)}$ is indeed equal to $1$. In conclusion, $\lim_{x\to 4}\dfrac{3x}{(x-2)(x+2)}=1$.